3.1 Some lemmas

The following result is natural; see [Reference Rippon and Stallard26, Lemma 1].

Lemma 3.1. Let f be a meromorphic function and let $\{E_n\}_{n=0}^{\infty }$ be a sequence of compact sets in $\mathbb {C}$ . If

$$ \begin{align*}E_{n+1}\subset f(E_n)\quad \text{for}\ n\geq 0,\end{align*} $$

then there exists $\xi $ such that $f^n(\xi )\in E_n$ for $n\geq 0$ . If $E_n\cap J(f)\not =\emptyset $ for $n\geq 0$ , then $\xi $ can be chosen to be in $J(f)$ .

The following result can be extracted from the proof of [Reference Rippon and Stallard26, Lemmas 6 and 7].

Lemma 3.2. Let f be a meromorphic function. If there exist two sequences $\{B_m\}_{m=0}^{\infty }$ and $\{V_m\}_{m=0}^{\infty }$ of compact sets with $\mathrm { dist}(0,B_m)\to \infty $ and $\mathrm {dist}(0,V_m)\to \infty $ as $m\to \infty $ , and a strictly increasing sequence of positive integers $\{m(k)\}$ such that

(3-1) $$ \begin{align} B_{m+1}\subseteq f(B_m),\ B_{m(k)-p(k)}\subseteq f(V_k),\ V_k\subseteq f(B_{m(k)}), \end{align} $$

where $0\leq p(k)\leq M$ for a fixed integer $M>0$ , then for any increasing sequence of positive numbers $\{a_n\}$ with $a_n\to \infty (n\to \infty )$ , there exists $\zeta \in I(f)$ such that for all sufficiently large n, $|f^n(\zeta )|\leq a_n$ . If, in addition, $B_m\cap J(f)\not =\emptyset \ \text { for all }\ m\geq 0$ , then we can require $\zeta \in J(f)$ .

Proof. We choose a subsequence $\{a_{n(m)}\}$ of $\{a_n\}$ such that

$$ \begin{align*}B_{p}\subset B(0,a_{n(m)})\quad \text{for}\ 0\leq p\leq m,\ \text{and}\ V_m\subset B(0,a_{n(m)}).\end{align*} $$

Inductively, we construct a sequence $\{s(k)\}$ of positive integers that is used to control the speed of iterates of f on $B_m$ . Set $d(k)=s(k)p(k)+2s(k)$ and $q(k)=d(0)+d(1)+d(2)+\cdots +d(k)=q(k-1)+d(k)$ .

Define $s(0)=0$ . Suppose that we have $s(k-1)$ , and so $d(k-1)$ and $q(k-1)$ are fixed. Take $s(k)$ such that

$$ \begin{align*}m(k)+q(k)>n(m(k+1)).\end{align*} $$

Let us construct a sequence $\{E_n\}$ of compact sets as follows:

$$ \begin{align*}\begin{array}{l} E_0=B_0,\ E_1=B_1,\ldots, E_{m(1)}=B_{m(1)},\\ E_{m(1)+1}=V_1,\\ E_{m(1)+2}=B_{m(1)-p(1)},\ldots,\ E_{m(1)+p(1)+2}=B_{m(1)},\\ ......\\ E_{m(1)+jp(1)+2j+1}=V_1,\\ E_{m(1)+jp(1)+2j+2}=B_{m(1)-p(1)},\ldots,\ E_{m(1)+(j+1)p(1)+2j+2}=B_{m(1)},\\ E_{m(1)+(j+1)p(1)+2(j+1)+1}=V_1,\ 0\leq j\leq s(1),\\ E_{m(1)+q(1)}=B_{m(1)},\ q(1)=d(1)=s(1)p(1)+2s(1),\\ E_{m(1)+1+q(1)}=B_{m(1)+1},\ldots,\ E_{m(2)+q(1)}=B_{m(2)},\\ E_{m(2)+q(1)+1}=V_2,\\ E_{m(2)+q(1)+2}=B_{m(2)-p(2)},\ldots,\ E_{m(2)+q(1)+p(2)+2}=B_{m(2)},\\ ...... \end{array}\end{align*} $$

that is to say, for $k\geq 0$ ,

$$ \begin{align*}E_n=\begin{cases}B_{n-q(k)}, & m(k)+q(k)\leq n\leq m(k+1)+q(k);\\ V_{k+1}, & n=m(k+1)+q(k)+jp(k+1)+2j+1;\\ \ & m(k+1)+q(k)+jp(k+1)+2j+2\\ B_{n-q(k)-(j+1)p(k+1)-2j-2}, & \leq n<m(k+1)+q(k)\\ \ &\quad +\,(j+1)p(k+1)+2j+1,\\ \ & 0\leq j\leq s(k+1).\end{cases}\end{align*} $$

Then it is easy to see that $E_{n+1}\subseteq f(E_n)$ . Since $m(k)-p(k)\to \infty \ (k\to \infty )$ and $\mathrm {dist}(0,E_n)\to \infty \ (n\to \infty )$ , in view of Lemma 3.1, there exists a point $\zeta \in B_0\cap I(f)$ such that $f^n(\zeta )\in E_n$ .

For $m(k)+q(k)\leq n\leq m(k+1)+q(k)$ ,

$$ \begin{align*}E_n=B_{n-q(k)}\subset B(0,a_{n(m(k+1))})\subset B(0,a_n)\end{align*} $$

by noting that $n(m(k+1))<m(k)+q(k)\leq n$ . When $n=m(k+1)+q(k)+jp(k+1)+2j+1$ , we have $n(k+1)<n(m(k+1))<m(k)+q(k)<n$ and so

$$ \begin{align*}E_n=V_{k+1}\subset B(0,a_{n(k+1)})\subset B(0,a_n).\end{align*} $$

For $m(k\kern1.7pt{+}\kern1.7pt1)\kern1.7pt{+}\kern1.7ptq(k)\kern1.7pt{+}\kern1.7pt jp(k\kern1.7pt{+}\kern1.7pt1)\kern1.7pt{+}\kern1.7pt 2j\kern1.7pt{+}\kern1.7pt2\leq n\kern1.7pt{<}\kern1.7pt m(k\kern1.7pt{+}\kern1.7pt1)\kern1.7pt{+}\kern1.7pt q(k)\kern1.7pt{+}\kern1.7pt(j\kern1.7pt{+}\kern1.7pt 1)p(k\kern1.7pt{+}\kern1.7pt1)\kern1.7pt{+}\kern1.7pt 2j + 1$ , that is, $m(k+1)-p(k+1)\leq n-q(k)-(j+1)p(k+1)-2j-2\leq m(k+1)$ ,

$$ \begin{align*}E_n=B_{n-q(k)-(j+1)p(k+1)-2j-2}\subset B(0,a_{n(m(k+1))})\subset B(0,a_n).\end{align*} $$

We have proved that for all $n\geq m(1)+q(1)$ , $E_n\subset B(0,a_n).$

The result in Lemma 3.2 also holds if the condition in (3-1) is replaced by

$$ \begin{align*}B_{m+1}\subseteq f(B_m),\ B_{m(k)-p(k)}\subseteq f(B_{m(k)}).\end{align*} $$

To be able to find sequences $\{B_m\}$ and $\{V_m\}$ in Lemma 3.2, the following covering theorem of annulus plays a key role. For a hyperbolic domain U, by $\lambda _U(z)$ , we denote the hyperbolic density of U at $z\in U$ and by $d_U(z_1,z_2)$ , the hyperbolic distance between $z_1$ and $z_2$ in U.

Lemma 3.3 [Reference Zheng34, Theorem 2.2].

Let f be analytic on a hyperbolic domain U with ${0\not \in f(U)}$ . If there exist two distinct points $z_1$ and $z_2$ in U such that $|f(z_1)|> e^{\kappa \delta }|f(z_2)|$ , where $\delta =d_U(z_1,z_2)$ and $\kappa =\Gamma (\tfrac 14)^4/(4\pi )^2=4.376\,879\,6\ldots ,$ then there exists a point $\hat {z}\in ~U$ such that $|f(z_2)|\leq |f(\hat {z})|\leq |f(z_1)|$ and

$$ \begin{align*} f(U)\supset A\bigg(e^{\kappa}\bigg(\frac{|f(z_2)|}{|f(z_1)|}\bigg)^{1/\delta}|f(\hat{z})|,\quad e^{-\kappa}\bigg(\frac{|f(z_1)|}{|f(z_2)|}\bigg)^{1/\delta}|f(\hat{z})|\bigg). \end{align*} $$

Moreover, if $|f(z_1)|\geq \exp ({\kappa \delta }/({1-\delta }))|f(z_2)|$ and $0<\delta <1$ , then

(3-2) $$ \begin{align} f(U)\supset A(|f(z_2)|,|f(z_1)|). \end{align} $$

In particular, for $\delta \leq \tfrac 16$ and $|f(z_1)|\geq e|f(z_2)|$ , we have (3-2).

We need a result on the existence of filling disks.

Lemma 3.4 (See [Reference Yang30, Lemma 3.4]).

Let f be a transcendental meromorphic function. Given arbitrarily $k>1,\ r$ , and R with $R>r$ satisfying

$$ \begin{align*}T(R,f)\geq \max\bigg\{240, \frac{240 \log(2R)}{\log k}, 12T(r,f), \frac{12T(kr,f)}{\log k} \log\frac{2R}{r}\bigg\},\end{align*} $$

then for large positive numbers q, there exists a point a with $r <|a|< 2R$ such that the disk

$$ \begin{align*}\Gamma: |z-a|<\frac{4\pi}{q}|a|\end{align*} $$

is a filling disk with index

(3-3) $$ \begin{align} m=c^*\frac{T(R,f)}{q^2(\log \frac{r}{R})^2},\end{align} $$

where $c^*> 0$ is an absolute constant.

There is $r_0\geq 0$ such that $T(r,f)\equiv $ constant for $r\in [0,r_0)$ and $T(r,f)$ is strictly increasing in $[r_0,+\infty )$ . Therefore, $T(r,f)$ is invertible in $[r_0,+\infty )$ . We denote the inverse of $s=T(r,f)$ in $[r_0, +\infty )$ by $r=T^{-1}(s,f)$ with $s\geq T(r_0,f)$ . For our purposes, we rewrite Lemma 3.4 as follows.

Lemma 3.5. Let f be a transcendental meromorphic function. Then there exists $R_0>0$ such that for $R>R_0$ and $q>R_0$ , the annulus $A(r, 3R)$ contains a filling disk

$$ \begin{align*} \Gamma: |z-a|<\frac{4\pi}{q}|a| \end{align*} $$

of f with index m given in (3-3) where

(3-4) $$ \begin{align} r=\frac{1}{2e}T^{-1}\bigg(\frac{T(R,f)}{12\log R},f\bigg). \end{align} $$

Since f is transcendental, it is clear that as $R\to \infty $ , we have $({T(R,f)}/{12\log R})\to \infty $ so that r in (3-4) goes to $\infty $ as $R\to \infty $ .

For the proof of Theorem 1.3, we need the following conclusion that asserts that a Borel direction decides the existence of a sequence of filling disks.

Lemma 3.6. Let f be a transcendental meromorphic function and $\arg z=\theta $ be a Borel direction of f of order $\mu $ with $0<\mu \leq \infty $ . Then there exists a sequence of filling disks:

$$ \begin{align*}\Gamma_1:=\{z: |z-z_j|<\varepsilon_j|z_j|\},\ z_j=|z_j|e^{i\theta},\ j=1,2,\ldots,\end{align*} $$

$$ \begin{align*}\lim_{j\to\infty}|z_j|=+\infty,\ \lim_{j\to\infty}\varepsilon_j=0\end{align*} $$

with index $m_j=|z_j|^{\mu _j}$ for a sequence $\mu _j$ such that $\mu>\mu _j\to \mu -0\ (j\to \infty ).$

Lemma 3.6 is essentially due to Rauch; see [Reference Yang30, Theorem 3.11]. What we mention is that the original conclusion of Rauch does not deal with the case of infinite order $\mu =\infty $ . However, we can get Lemma 3.6 without difficulty by making a small change of the proof of [Reference Yang30, Theorem 3.11].

Now we recall the Ahlfors–Shimizu characteristic of a meromorphic function; see [Reference Hayman17]. By $f^{\#}(z)$ , we mean the spherical derivative of f at z. For a closed domain D, define

$$ \begin{align*}\mathcal{A}(D,f)=\iint_D(f^{\#}(z))^2\,{d}\sigma(z),\end{align*} $$

where $\sigma (z)$ is the area element and write $\mathcal {A}(r,f)$ for $\mathcal {A}(\overline {B}(0,r),f)$ . The Ahlfors–Shimizu characteristic of f is defined as

$$ \begin{align*}\mathcal{T}(r,f)=\int_0^r\frac{\mathcal{A}(t,f)}{t}\,{d}t.\end{align*} $$

Then,

$$ \begin{align*}|T(r,f)-\mathcal{T}(r,f)-\log^+|f(0)||\leq \tfrac{1}{2}\log 2.\end{align*} $$

To confirm the existence of filling disks in view of Lemmas 3.4 and 3.5 in our proofs of theorems, we need to give a convenient version of (1-6).

Lemma 3.7. Let f be a transcendental meromorphic function. Assume that for a ${0<c<1}$ and all sufficiently large r, (1-6) holds. Then for large r,

(3-5) $$ \begin{align} \frac{T(r^2,f)}{\log r^2}>12~T( er,f). \end{align} $$

Proof. Using the condition in (1-6) repeatedly, for $\sigma>0$ ,

(3-6) $$ \begin{align} T(r^{1+\sigma},f)&=T(e^{\sigma\log r}r,f)>\bigg(1+\frac{1}{((1+\sigma)\log r-1)^c}\bigg)T(e^{(1+\sigma)\log r-1},f)\nonumber\\ &>\prod_{k=1}^{[\sigma\log r-2]}\bigg(1+\frac{1}{((1+\sigma)\log r-k)^c}\bigg)\cdot T(e^2r,f)\nonumber\\ &>\bigg(1+\frac{1}{((1+\sigma)\log r)^c}\bigg)^{[\sigma\log r-2]}T(e^2r,f)\nonumber\\ &>\exp\frac{[\sigma\log r-2]}{((1+\sigma)\log r)^c+1}\cdot T(e^2r,f)\nonumber\\ &>12(1+\sigma)(\log r) T(e^2r,f)\nonumber\\ &>12\ T(e^2r,f),\end{align} $$

where $[x]$ denotes the integer part of x.

That is to say, when $\sigma $ is chosen to be $1$ , we obtain (3-5) since $T(r,f)$ is increasing on r.

Let us make a remark on (3-5). For an arbitrarily given large integer $N>0$ and sufficiently large r, we have from (3-5) that

$$ \begin{align*}T(r,f)\geq \frac{6^N}{2^{2^N-1}}(\log r)^NT(r^{1/2^N},f).\end{align*} $$

Therefore,

(3-7) $$ \begin{align}\lim_{\overline{r\to\infty}}\frac{T(r,f)}{(\log r)^N}=\infty.\end{align} $$

The final lemma can be deduced using calculus.

Lemma 3.8. For $R>0$ , the function $\sqrt {({1+x^2})/({1+(x-R)^2})}$ is increasing in $[0,\ \tfrac 12(R+\sqrt {4+R^2})]$ and decreasing in $[\tfrac 12(R+\sqrt {4+R^2}),\infty )$ and

$$ \begin{align*}\sqrt{\frac{1+x^2}{1+(x-R)^2}}\leq \frac{1}{2}\big(R+\sqrt{4+R^2}\big)\quad \text{ for all }\ x>0.\end{align*} $$

3.2 Proof of Theorem 1.1

Let $R_0$ be as in Lemma 3.5. Take $R_1$ with $q=\log \log R_1>R_0$ ; in view of Lemma 3.5, there exists a filling disk $\Gamma _1$ in $A(r_1,3R_1)$ with index

$$ \begin{align*}m_1=c^*\frac{T(R_1,f)}{(\log\log R_1)^2(\log R_1)^2}>6\log R_1\end{align*} $$

and $r_1$ defined by (3-4) with $R=R_1$ . Note that $\chi (z,\infty )>\tfrac 32e^{-m_1}$ on $|z|=\tfrac 12e^{m_1}$ and the spherical diameter of the circle $|z|=\tfrac 12e^{m_1}$ is larger than $2e^{-m_1}$ . According to the definition of filling disks, there exists a point $z_1\in \Gamma _1$ such that $|f(z_1)|=\tfrac 12e^{m_1}$ . Set

$$ \begin{align*} W_1=\{z: 2<|z|<\tfrac{1}{128}|f(z_1)|\}. \end{align*} $$

We need to treat three cases.

Case A. Assume that there exists a point $w_0\in W_1$ such that $w_0\not \in f(5\Gamma _1)$ and f is analytic in $5\Gamma _1$ . Here and henceforth, for a disk $\Gamma =B(a,r)$ , we define $5\Gamma =B(a,5r)$ . Since $\mathrm {diam}_{\chi }(B(w_0,2))>5e^{-m_1}$ , there exists a point $z_0\in \Gamma _1$ such that $|f(z_0)-w_0|\leq 2$ . Set $g(z)=f(z+w_0)-w_0$ . Then $0\not \in g(H_1)$ with $H_1=5\Gamma _1-w_0$ . In view of Lemma 3.3, by noting that $d_{H_1}(z_0-w_0,z_1-w_0)=d_{5\Gamma _1}(z_0,z_1)\leq \log {17}/{7}<1$ and $|g(z_1-w_0)|\geq |f(z_1)|-|w_0|\geq \tfrac 12|f(z_1)|,$

$$ \begin{align*}g(H_1)\supset A(2,|g(z_1-w_0)|)\supset A\big(2,\tfrac{1}{2}|f(z_1)|\big).\end{align*} $$

Set $\hat {R}_2={1}/{32}|f(z_1)|$ . In view of Lemma 3.5, g has a filling disk $\hat {\Gamma }_2$ in $A(\hat {r}_2,3\hat {R}_2)$ with index

$$ \begin{align*} \hat{m}_2=c^*\frac{T(\hat{R}_2,g)}{(\log\log \hat{R}_2)^2(\log \hat{R}_2)^2} \end{align*} $$

and $\hat {r}_2={1}/{2e}T^{-1}({T(\hat {R}_2,g)}/{12\log \hat {R}_2},g)>32.$ Thus, $5\hat {\Gamma }_2\subset g(H_1).$

By $\hat {\gamma }_{21}$ and $\hat {\gamma }_{22}$ , we denote the two exceptional disks of g for the filling disk $\hat {\Gamma }_2$ . Then $\gamma _{21}=\hat {\gamma }_{21}+w_0$ and $\gamma _{22}=\hat {\gamma }_{22}+w_0$ are the exceptional disks of f for $\Gamma _2=\hat {\Gamma }_2+w_0$ . We can assume without any loss of generality that the center of $\gamma _{21}$ is a finite number and $\infty $ is the center of $\gamma _{22}$ . For any two points $z_1$ and $z_2$ in $\gamma _{21}$ , set $\hat {z}_1=z_1-w_0$ and $\hat {z}_2=z_2-w_0$ in $\hat {\gamma }_{21}$ . Then

$$ \begin{align*} \chi(z_1,z_2) &=\frac{\sqrt{1+|\hat{z}_1|^2} \sqrt{1+|\hat{z}_2|^2}}{\sqrt{1+|z_1|^2}\sqrt{1+|z_2|^2}}\chi(\hat{z}_1,\hat{z}_2)\\[5pt] &\leq \sqrt{\frac{1+|\hat{z}_1|^2}{1+(|\hat{z}_1|-|w_0|)^2}} \sqrt{\frac{1+|\hat{z}_2|^2}{1+(|\hat{z}_2|-|w_0|)^2}} \chi(\hat{z}_1,\hat{z}_2)\\[5pt] &\leq\frac{1}{4}(|w_0|+\sqrt{4+|w_0|^2})^2\chi(\hat{z}_1, \hat{z}_2)<3R_2^2\chi(\hat{z}_1,\hat{z}_2). \end{align*} $$

Additionally, for a point $\hat {z}\in \hat {\gamma }_{22}$ , by noting that $\chi (\hat {z},\infty )<e^{-\hat {m}_2}$ , we have $|\hat {z}|>e^{\hat {m}_2}-1>R^3_2>2|w_0|$ and so for $z=\hat {z}+w_0\in \gamma _{22}$ ,

$$ \begin{align*} \chi(z,\infty)&=\frac{\sqrt{1+|\hat{z}|^2}}{\sqrt{1+|z|^2}}\chi(\hat{z},\infty)\leq \sqrt{\frac{1+|\hat{z}|^2}{1+(|\hat{z}|-|w_0|)^2}}\,\chi(\hat{z},\infty)\\ &<\frac{|\hat{z}|}{|\hat{z}|-|w_0|}\chi(\hat{z},\infty)<2\chi(\hat{z},\infty). \end{align*} $$

Set

(3-8) $$ \begin{align} m_2=c^*\frac{T(R_2,f)}{5(\log\log R_2)^2(\log R_2)^3}, \end{align} $$

with $R_2=\tfrac 14\hat {R}_2.$ Since $B(0,R_2)\subset B(0,\tfrac 12\hat {R}_2)+w_0$ ,

$$ \begin{align*} \mathcal{A}\big(\tfrac{1}{2}\hat{R}_2,f(z+w_0)\big) =\mathcal{A}\big(B\big(0,\tfrac{1}{2}\hat{R}_2\big)+w_0,f(z)\big)\end{align*} $$

$$ \begin{align*} \geq \mathcal{A}(B(0,R_2),f(z))=\mathcal{A}(R_2,f).\end{align*} $$

Therefore,

$$ \begin{align*} T(\hat{R}_2,g)&=T(\hat{R}_2,f(z+w_0)-w_0)\\ &=N(\hat{R}_2,f(z+w_0))+m(\hat{R}_2,f(z+w_0)-w_0)\\ &\geq T(\hat{R}_2,f(z+w_0))-\log |w_0|-\log 2\\ &\geq T(\hat{R}_2,f(z+w_0))-T\bigg(\frac{1}{2}\hat{R}_2,f(z+w_0)\bigg)-\log |w_0|-\log 2\\ &\geq \mathcal{T}(\hat{R}_2,f(z+w_0))-\mathcal{T} \bigg(\frac{1}{2}\hat{R}_2,f(z+w_0)\bigg)-\log|w_0|-\frac{3}{2}\log 2\\ &= \int_{{1}/{2}\hat{R}_2}^{\hat{R}_2}\frac{\mathcal{A}(t,f(z+w_0))}{t}\,{d}t-\log|w_0|-\frac{3}{2}\log 2\\ &\geq \mathcal{A}\bigg(\frac{1}{2}\hat{R}_2,f(z+w_0)\bigg)\log 2-\log|w_0|-\frac{3}{2}\log 2\\ &\geq \mathcal{A}(R_2,f)\log 2-\log|w_0|-\frac{3}{2}\log 2\\ & \geq \frac{\log 2}{\log R_2}\int_{1}^{R_2}\frac{\mathcal{A}(t,f)}{t}\,{d}t-\log|w_0|-\frac{3}{2}\log 2\\ &=\frac{\log 2}{\log R_2}T(R_2,f)+O(1)-\log|w_0|-\frac{3}{2}\log 2\\ &\geq \frac{1}{2\log R_2}T(R_2,f). \end{align*} $$

This implies that $\hat {m}_2>2m_2$ and so $3R^2_2e^{-\hat {m}_2}<3R^2_2e^{-2{m}_2}<e^{-m_2}.$ Therefore, f has the filling disk $\Gamma _2$ with index $m_2.$ Since $5\hat {\Gamma }_2\subset g(H_1)$ , we have $5\Gamma _2-w_0\subset f(5\Gamma _1)-w_0$ and so $5\Gamma _2\subset f(5\Gamma _1).$

Case B. Assume that $W_1\subset f(5\Gamma _1)$ and f is analytic in $5\Gamma _1$ . In view of Lemma 3.5, we have a filling disk $\Gamma _2$ of f in $A(r_2,3R_2)\subset W_1$ with index $m_2$ , where $R_2={1}/{640}|f(z_1)|$ , and $5\Gamma _2\subset f(5\Gamma _1)$ .

Case C. Assume that $5\Gamma _1$ contains a pole of f. Then for some $\hat {r}_2>0$ , $\{z:\ |z|>\hat {r}_2\}\subset f(5\Gamma _1)$ . Take a sufficiently large $R_2>R^3_1$ such that $r_2>5\hat {r}_2$ . In view of Lemma 3.5, there exists a filling disk $\Gamma _2$ of f in $A(r_2,3R_2)$ with index $m_2$ . Obviously, $5\Gamma _2\subset A(r_2/5,15R_2)\subset f(5\Gamma _1).$

In one word, there exists a filling disk $\Gamma _2$ of f with index $m_2$ given in (3-8) where $R_2>R^3_1$ and $5\Gamma _2\subset f(5\Gamma _1)$ . Proceeding step by step, we obtain a sequence of filling disks $\{\Gamma _n\}$ of f with index $m_n$ given by (3-8) with $R_2$ replaced by $R_n$ , and $R_n>R_{n-1}^3\to \infty\ (n\to \infty )$ and $5\Gamma _n\subset f(5\Gamma _{n-1})$ . Since f is meromorphic on the complex plane, f takes any value at most finitely many times on any bounded subset of the complex plane and therefore, since $m_n\to \infty\ (n\to \infty )$ , we know that $\mathrm {dist}(\Gamma _n,0)\to \infty\ (n\to \infty ).$ It is easily seen that for every n, $\Gamma _n\cap J(f)\not =\emptyset .$ In view of Lemma 3.1, there exists a point $a\in I(f)\cap J(f)$ such that $f^n(a)\in 5\Gamma _n$ . Of course, $5\Gamma _n$ is also a filling disk of f with index  $m_n$ .

We complete the proof of Theorem 1.1.

3.3 Proof of Theorem 1.3

From the proof of Theorem 1.1, we have a sequence of filling disks $B_n$ such that $B_{n+1}\subset f(B_n)$ centered at $z_n\to \infty $ with index $m_n$ , and having order greater than or equal to $\lambda $ . Therefore, the spherical radius of the exceptional disks is $e^{-m_n}\to 0\ (n\to \infty )$ .

Noting that $B_{\mathrm {direct}}(f)$ is closed in $[0,2\pi ]$ , we can choose a sequence of $\{\theta _p\}_{p=1}^N$ with $1\leq N\leq +\infty $ such that the closure $\overline {\{\theta _p:\ p=1,2,\ldots , N\}}=B_{\mathrm {direct}}(f)$ . In view of Lemma 3.6, for every $\theta _p$ , there exists a sequence of filling disks $\{A_{pj}\}_{j=1}^{\infty }$ with index $m_{pj}$ centered at $z_{pj}=r_{pj}e^{i\theta _p}, j=1,2,\ldots $ with $r_{pj}\to \infty\ (j\to \infty )$ and $m_{p(j+1)}>m_{pj}$ . Additionally, we can require that $r_{p(j+1)}>2r_{pj}$ and $r_{(p+1)1}>2r_{p1}\to +\infty\ (p\to \infty )$ . Since the exceptional disks of $A_{pj}$ have spherical radius at most $e^{-m_{pj}}$ , choosing sufficiently large $r_{p1}$ , we have that for every j, there exists a $B_{n(pj)}$ with ${n(pj)\to \infty\ (p\to \infty )}$ such that

$$ \begin{align*}B_{n(pj)}\subset f(A_{pj}).\end{align*} $$

Now let us write $A_{pj},\ j=1,2,\ldots , p=1,2,\ldots $ in the following order:

$$ \begin{align*}A_{11},\ A_{12},\ A_{21}, A_{13}, A_{22}, A_{31},\ldots.\end{align*} $$

Take a sufficiently large $n_0$ and then one of $A_{11},\ A_{12},\ A_{13}$ is in $f(B_{n_0})$ , and denote it by $C_{11}$ . Then $f(C_{11})$ contains one $B_{n_{11}}$ . Take $B_{n_{11}+1},\ldots ,B_{n_{11}+p_{11}}$ such that $f(B_{n_{11}+p_{11}})$ contains one of $A_{12},\ A_{13}, A_{14}$ , and denote it by $C_{12}$ . Then $f(C_{12})$ contains one $B_{n_2}$ . Take $B_{n_{12}+1},\ldots ,B_{n_{12}+p_{12}}$ such that $f(B_{n_{12}+p_{12}})$ contains one of $A_{21}, A_{22}, A_{23}$ , and denote it by $C_{21}$ . We go on forever in this way to obtain a sequence of filling disks:

where $C_{sk}$ is one of $A_{sk}, A_{s(k+1)}$ , and $A_{s(k+2})$ . In view of Lemma 3.1, there exists a point $a\in I(f)\cap J(f)$ such that $f^n(a)$ goes along the sequence of the above filling disks. Then a satisfies our requirement.

3.4 Proof of Theorem 1.4

Under (1-6), we have (3-5). Take $r_1$ sufficiently large and set $R_1=r_1^2$ such that

$$ \begin{align*}T(R_1,f)\geq \max\bigg\{240, \frac{240 \log(2R_1)}{\log 2}, 12T(er_1,f)\log (2r_1)\bigg\},\end{align*} $$

$q_1=\log r_1$ and, in view of (3-7),

$$ \begin{align*}c^*\frac{T(R_1,f)}{(\log r_1)^4}>2\log (1+2R_1),\end{align*} $$

where $c^*$ is the constant in Lemma 3.4. Applying Lemma 3.4, there exists a $z_1$ lying in annulus $\{z: r_1<|z|<2R_1\}$ such that

$$ \begin{align*}\Gamma_1 := \bigg\{ z : |z - z_1| < \frac{4\pi}{q_1}|z_1|\bigg\}\end{align*} $$

is a filling disk of f with index

$$ \begin{align*}n_1=c^*\frac{T(R_1,f)}{q_1^2(\log r_1)^2}=c^*\frac{T(R_1,f)}{(\log r_1)^4}.\end{align*} $$

Take $r_2,\ R_2=r_2^2$ and $q_2=\log r_2$ such that

$$ \begin{align*}r_2>\frac{2R_1+(1+2R_1)e^{-n_1}}{1-e^{-n_1}(1+2R_1)}.\end{align*} $$

This implies that $\chi (r_2,2R_1)>e^{-n_1}$ , where $\chi (z,w)$ denotes the spherical distance between z and w. There exists a $z_2$ lying in the annulus $\{z: r_2<|z|<2R_2\}$ such that

$$ \begin{align*}\Gamma_2 := \bigg\{ z : |z - z_2| < \frac{4\pi}{q_2}|z_2|\bigg\}\end{align*} $$

is a filling disk of f with index

$$ \begin{align*}n_2=c^*\frac{T(R_2,f)}{(\log r_2)^4}.\end{align*} $$

Additionally, the spherical distance between $\Gamma _1$ and $\Gamma _2$ is at least $\chi (r_2,2R_1)>e^{-n_1} >e^{-n_2}.$

Using the same method, we have $r_j, R_j, q_j, n_j$ and $\Gamma _j\ (j=1,2,\ldots ,5)$ such that the spherical distance between $\Gamma _j$ and $\Gamma _i$ with $i\not =j$ is larger than $e^{-n_t}\ (t=1,2,\ldots ,5)$ .

Take $B_1=\Gamma _1$ . We can have $B_2=\Gamma _i$ for some $2\leq i\leq 4$ with $f(B_1)\supset B_2$ and then $B_3=\Gamma _j$ for some $j\in \{2,3,4,5\}\setminus \{i\}$ with $f(B_2)\supset B_3$ . It is easy to see that $f(B_3)\supset B_1, B_2$ or $B_3$ . Starting from $B_3$ , we apply the same method to obtain $B_4$ and $B_5$ such that $f(B_j)\supset B_{j+1}$ with $j=3,4$ and $f(B_5)\supset B_3, B_4$ , or $B_5$ . Thus, we obtain a sequence of disks $\{B_n\}$ such that $f(B_n)\supset B_{n+1}$ and $f(B_{2n+1})\supset B_{2n-1}, B_{2n}$ , or $B_{2n+1}$ .

We can take $r_n$ such that $\lim \nolimits _{n\to \infty }({\log T(R_n,f)}/{\log R_n})=\rho (f)$ . Applying Lemma 3.2, we complete the proof of Theorem 1.4.

3.5 Proof of Theorem 1.5

It follows from (3-6) that

$$ \begin{align*}\frac{1}{2e}T^{-1}\bigg(\frac{T(r^{1+\sigma},f)}{12\log r^{1+\sigma}},f\bigg)>r.\end{align*} $$

In view of Lemma 3.5, for any $\sigma>0$ and all sufficiently large r, the annulus $A(r^{1-\sigma },r)$ contains a filling disk with index $m(r)\to \infty\ (r\to \infty ).$ Since $J(f)$ is nonempty and unbounded, $F(f)$ cannot contain any filling disks with large index so that given arbitrarily $0<\sigma <1$ , for large r, under (1-6), $F(f)$ cannot contain any annulus $A(r^{1-\sigma },r)$ .

Suppose that f has a multiply connected Fatou component. Then there exists a sequence of annuli $\{A(r_n^{1-\sigma },r_n)\}$ with $r_n\to \infty $ for some $0<\sigma <1$ in the Fatou set $F(f)$ ; see [Reference Bergweiler, Rippon and Stallard9, Theorem 1.2] and [Reference Zheng34, Theorem 1.1]. This derives a contradiction. Theorem 1.5 follows.

4.1 Some lemmas

We need the Nevanlinna characteristic in an angle; see [Reference Goldberg and Ostrovskii16, Reference Zheng33]. We set

$$ \begin{align*} \Omega(\alpha,\beta)=\{z: \alpha<\arg z<\beta\} \end{align*} $$

with $0\leq \alpha <\beta \leq 2\pi $ and denote by $\overline {\Omega }(\alpha ,\beta )$ the closure of $\Omega (\alpha ,\beta )$ . Let $f(z)$ be meromorphic on the angle $\overline {\Omega }(\alpha ,\beta )$ . We define

$$ \begin{align*} A_{\alpha,\beta}(r,f)&= \frac{\omega}{\pi}\int_1^r \bigg(\frac{1}{t^{\omega}}-\frac{t^{\omega}}{r^{2\omega}}\bigg) \{\log^+|f(te^{i\alpha})|+\log^+|f(te^{i\beta})|\}\frac{{d}t}{t};\\B_{\alpha,\beta}(r,f)&= \frac{2\omega}{\pi r^w}\int_{\alpha}^{\beta}\log^+|f(re^{i\theta})|\sin \omega(\theta-\alpha)\,{d}\theta;\\C_{\alpha,\beta}(r,f)&= 2\sum_{1<|b_n|<r}\bigg(\frac{1}{|b_n|^{\omega}} -\frac{|b_n|^{\omega}}{r^{2\omega}}\bigg)\sin \omega(\beta_n-\alpha), \end{align*} $$

where $\omega =\pi /(\beta -\alpha )$ and $b_n=|b_n|e^{i\beta _n}$ are poles of $f(z)$ in $\overline {\Omega }(\alpha ,\beta )$ appearing according to their multiplicities. Additionally, define $\overline {C}_{\alpha ,\beta }(r,f)$ in the same form as $C_{\alpha ,\beta }(r,f)$ for distinct poles $b_n$ of $f(z)$ , that is, ignoring their multiplicities. For $a\in \mathbb {C}$ , we write $C_{\alpha ,\beta }(r,f=a)$ for $C_{\alpha ,\beta }(r,1/(f-a))$ . The Nevanlinna angular characteristic is defined as

$$ \begin{align*}S_{\alpha,\beta}(r,f)=A_{\alpha,\beta}(r,f)+B_{\alpha,\beta}(r,f)+C_{\alpha,\beta}(r,f).\end{align*} $$

Lemma 4.1 [Reference Zheng33, Lemma 2.2.2].

Let $f(z)$ be a meromorphic function on $\overline {\Omega }(\alpha , \beta )$ . Then we have the following:

$$ \begin{align*}C_{\alpha,\beta}(r,f=a)\leq 4\omega \frac{N(r,\Omega,f=a)}{r^{\omega}} +2\omega^2 \int_{1}^{r}\frac{N(t,\Omega,f=a)}{t^{\omega+1}}\,{d}t.\end{align*} $$

The inequality also holds for $a=\infty $ .

Lemma 4.2 [Reference Zheng33, Equation (2.2.6) and Lemma 2.5.3].

Let $f(z)$ be a meromorphic function. Then for any two distinct values $a_1$ and $a_2$ on $\mathbb {C}$ ,

$$ \begin{align*}S_{\alpha,\beta}(r,f)\leq C_{\alpha,\beta}(r,f)+\sum_{\nu=1}^2 \overline{C}_{\alpha,\beta}(r,f=a_v)+O(\log rT(r,f))\end{align*} $$

for all $r>0$ with the possible exception of a finite-measure set of r.

We need the following lemma, which is established in terms of the hyperbolic metric.

Lemma 4.3 [Reference Zheng34, Theorem 2.4].

Let $h(z)$ be an analytic function on the annulus $A(r,R)=\{z: r<|z|<R\}$ with $0<r<R<\infty $ such that $|h(z)|>1$ on $A(r,R)$ . Then

$$ \begin{align*} \log L(\rho,h) \geq \exp\bigg(-\frac{\pi^2}{2}\max\bigg\{\frac{1}{\log \frac{R}{\rho}},\frac{1}{\log \frac{\rho}{r}}\bigg\}\bigg)\log M(\rho,h), \end{align*} $$

where $\rho \in (r,R)$ and $L(\rho ,h)=\min \{|h(z)|: |z|=\rho \}$ .

In Lemma 4.3, when $\rho =\sqrt {rR}$ ,

$$ \begin{align*} \log M(\rho,h) \leq \exp\bigg(\frac{\pi^2}{\log \frac{R}{r}}\bigg)\log L(\rho,h). \end{align*} $$

4.2 Proof of Theorem 1.7

In contrast, suppose that there exists a sequence of annuli $A_n=\{z:r_n<|z|<R_n\}$ in $F(f)$ with $R_n\geq (1+\phi (r_n)/{\log}\, T(r_n,f))r_n$ , $r_{n+1}>r_n$ and $r_n\to \infty\ (n\to \infty )$ . We treat two cases.

Case A. There exists a subsequence of $A_n$ such that $f(A_n)\subset \{|z|>1\}$ . Without loss of generality, we assume that for all n, $f(A_n)\subset \{|z|>1\}$ . Set $\rho _n=\sqrt {R_nr_n}.$ Since $|f(z)|>1$ on $A_n$ , using Lemma 4.3, we have

$$ \begin{align*}\log L(\rho_n,f)\geq \lambda_n\log M(\rho_n,f),\end{align*} $$

where $\lambda _n=\exp (-{\pi ^2}/{\log ({R_n}/{r_n})})$ .

For the angular domain $\Omega (\alpha ,\beta )$ , according to the definition of $B_{\alpha ,\beta }(\rho _n,f)$ ,

(4-1) $$ \begin{align} \rho_n^{\omega}B_{\alpha, \beta}(\rho_n,f) & = \frac{2\omega}{\pi}\int_{\alpha}^{\beta} \log^{+}|f(\rho_n e^{i\phi})|\sin (\omega(\phi-\alpha))\,{d}\phi\notag\\ & \geq \frac{2\omega}{\pi}\log L(\rho_n,f)\int_{\alpha}^{\beta} \sin (\omega(\phi-\alpha))\,{d}\phi\notag\\ & =\frac{4}{\pi}\log L(\rho_n,f)\geq\frac{4\lambda_n}{\pi}\log M(\rho_n,f). \end{align} $$

However, for any two distinct complex numbers $a_v ~(v=1,2)$ , we use Lemmas 4.1 and 4.2 in turn to obtain that

(4-2) $$ \begin{align} \rho_n^{\omega}B_{\alpha, \beta}(\rho_n,f)&\leq \rho_n^{\omega}(C_{\alpha, \beta}(\rho_n,f=a_1) +C_{\alpha, \beta}(\rho_n,f=a_2)) +O(\rho_n^{\omega}\log \rho_nT(\rho_n,f))\nonumber\\ & \leq 4\omega N(\rho_n)+2\omega^2 \rho_n^{\omega} \int_1^{\rho_n}\frac{N(t)}{t^{\omega+1}}\,dt+O(\rho_n^{\omega} \log \rho_nT(\rho_n,f))\nonumber\\ & \leq 4\omega N(\rho_n)+2\omega \rho_n^{\omega}N(\rho_n)+O(\rho_n^{\omega}\log \rho_nT(\rho_n,f))\nonumber\\ & = (4\omega +2\omega \rho_n^{\omega})N(\rho_n)+O(\rho_n^{\omega}\log \rho_nT(\rho_n,f)), \end{align} $$

where $N(\rho _n)=N(\rho _n,\Omega ,f=a_1)+N(\rho _n,\Omega ,f=a_2)$ . Combining (4-1), (4-2), and (1-7) yields

$$ \begin{align*}(4\omega +2\omega \rho_n^{\omega})N(\rho_n)+O(\rho_n^{\omega}\log \rho_nT(\rho_n,f)) \geq \lambda_n\log M(\rho_n,f))\geq K\lambda_nT(\rho_n,f)\end{align*} $$

for some positive constant K. Thus,

(4-3) $$ \begin{align} K\lambda_nT(\rho_n,f)\leq 2\max\{(4\omega +2\omega \rho_n^{\omega})N(\rho_n),O(\rho_n^{\omega}\log \rho_nT(\rho_n,f))\}.\end{align} $$

It follows from the definition of lower order that

$$ \begin{align*}\lim\limits_{n\to\infty} \frac{\log T(\rho_n,f)}{\log \rho_n}\geq\lambda\end{align*} $$

for the above sequence $\{\rho _n\}$ . By noting that

$$ \begin{align*} \log\lambda_n&=-\frac{\pi^2}{\log(R_n/r_n)}\geq-\frac{\pi^2}{\log(1+\phi(r_n)/{\log}\, T(r_n,f))}\\ &\sim-\frac{\pi^2}{\phi(r_n)}\log T(r_n,f)\geq -\frac{\pi^2}{\phi(r_n)}\log T(\rho_n,f), \end{align*} $$

we have

$$ \begin{align*} \lim\limits_{\overline{n\to\infty}}\frac{\log \lambda_n}{\log T(\rho_n,f)}\geq 0. \end{align*} $$

In view of (4-3),

$$ \begin{align*} \lim_{n\rightarrow \infty}\frac{\log N(\rho_n)}{\log T(\rho_n,f)}\geq 1-\frac{\omega}{\lambda}. \end{align*} $$

Then we get a contradiction for a and b.

Case B. Set

$$ \begin{align*}c_n=1+\frac{\phi(r_n)}{12\log T(r_n,f)}.\end{align*} $$

When $c_n\leq 2$ , we have $c_n^3\leq 1+\phi (r_n)/{\log}\, T(r_n,f).$ Consider the annulus ${B_n=A(r_n,c_n^3r_n)\subset A_n}$ and $C_n=A(c_nr_n,c_n^2r_n)$ . In view of the implication in Case A, we can assume that for all n, $f(C_n)\cap \{|z|\leq 1\}\not =\emptyset $ . Then there exist two points $z_0\in C_n$ and $z_n$ with $|z_n|=\rho _n=c^{3/2}_nr_n$ , in view of (1-7), such that

$$ \begin{align*}|f(z_0)|= 1+|c|\quad \text{and}\quad \log |f(z_n)|\geq KT(\rho_n,f)>\log\rho_n\quad \text{for } n\geq N,\end{align*} $$

where $c\in J(f)$ with $|c|=\min \nolimits _{z\in J(f)}|z|$ , K is a positive number, and N is a positive integer. Thus, for $n\geq N$ and $N\leq k\leq n$ , $A_k\cap f(A_n)\not =\emptyset .$ This implies that $\bigcup _{n=N}^{\infty } A_n\subset ~U$ , where U is a Fatou component of f with $f(U)\subseteq U$ .

A simple calculation yields

$$ \begin{align*}\lambda_{B_n}(z)\leq\frac{2\sqrt{3}\pi}{9\log c_n}\frac{1}{|z|}, \quad z\in C_n\quad \text{and}\quad \lambda_{B_n}(z)=\frac{\pi}{3\log c_n}\frac{1}{|z|},\ |z|=\rho_n.\end{align*} $$

Then

$$ \begin{align*}d_{B_n}(z_0,z_n)\leq \int_{c_nr_n}^{\rho_n}\frac{2\sqrt{3}\pi}{9\log c_n}\frac{{d}t}{t}+\int_0^{\pi}\frac{\pi}{3\log c_n}\,{d}\theta \leq \frac{\sqrt{3}\pi}{9}+\frac{\pi^2}{3\log c_n}\end{align*} $$

and

$$ \begin{align*}\frac{1}{\delta_n}\geq\frac{9\log c_n}{\sqrt{3}\pi\log c_n+3\pi^2}\geq\frac{\log c_n}{4+\log c_n},\end{align*} $$

where $\delta _n=d_{B_n}(z_0,z_n)$ . It follows that for sufficiently large n,

$$ \begin{align*}|f(z_n)|\geq \exp(KT(\rho_n,f))> e^{\kappa\delta_n}(1+2|c|)+|c|= e^{\kappa\delta_n}(|f(z_0)|+|c|)+|c|.\end{align*} $$

Set $g(z)=f(z)-c$ . Then $0\not \in g(B_n)$ and $|g(z_n)|\geq e^{\kappa \delta _n}|g(z_0)|$ . In view of Lemma 3.3, we have $g(B_n)\supseteq A(d^{-1}_nt_n, d_nt_n)$ with $t_n\geq |g(z_0)|\geq 1$ and

$$ \begin{align*}d_n:=e^{-\kappa}\bigg(\frac{|g(z_n)|}{|g(z_0)|}\bigg)^{1/\delta_n}\geq \exp\bigg(-\kappa+\frac{\log c_n}{4+\log c_n}\frac{K}{2}T(\rho_n,f)\bigg)>\rho_n^5.\end{align*} $$

Thus,

$$ \begin{align*}f(B_n)\supseteq A(d^{-1}_nt_n, d_nt_n)+q\supseteq A(d^{-1}_nt_n+|c|, d_nt_n-|c|).\end{align*} $$

Set $s_n=\rho _nt_n\geq \rho _n$ . Then

$$ \begin{align*}A(s_n, \rho_n^3s_n)\subset A(d^{-1}_nt_n+|c|, d_nt_n-|c|).\end{align*} $$

This implies that the Fatou component U contains a sequence of annuli ${D_n = A(s_n, \rho _n^3s_n)}$ .

For a simple statement, we assume without loss of generality that $c=0$ . We can assume that $f(E_n)\cap \{z:\ |z|\leq 1\}\not =\emptyset $ with $E_n=A(\rho _ns_n, \rho _n^2s_n)$ . We can find two points $z_0'\in E_n$ and $z_n'$ with $|z_n'|=\rho _n^{3/2}s_n$ such that

$$ \begin{align*}|f(z_0')|=1\quad \text{and}\quad \log |f(z_n')|\geq KT(\rho_n^{3/2}s_n,f)>2\log r_n.\end{align*} $$

Additionally,

$$ \begin{align*}d_{D_n}(z_0',z_n')\leq\frac{\sqrt{3}\pi}{9}+\frac{\pi^2}{3\log \rho_n}<\frac{9}{10}.\end{align*} $$

Therefore, in view of Lemma 3.3, $f(D_n)\supset A(|f(z_0')|,f(z_n')|)\supset A(1, r_n^2)$ and furthermore, $A(1,r_n^2)\subset U$ and so $A(1,R_n)\subset U$ . This implies that $U\supset \{z: |z|>1\}$ . A contradiction is derived.

From Cases A and B, the Fatou set $F(f)$ contains no annuli mentioned in Theorem 1.7.

4.3 Proof of Theorem 1.8

Consider the meromorphic function with the following form:

$$ \begin{align*}f(z)=cz+d+\sum_{n=1}^{\infty} c_n\bigg(\frac{1}{a_n-z}-\frac{1}{a_n}\bigg), \end{align*} $$

where $c, d, c_n$ , and $a_n$ are real numbers with $a_n\to \infty (n\to \infty )$ , $c\geq 0$ , and $c_n>0$ such that

$$ \begin{align*} \sum_{n=1}^{\infty}\frac{c_n}{a_n^2}<+\infty. \end{align*} $$

For such a function f, the real axis is completely invariant under f and so $J(f)$ is completely on the real axis, see [Reference Baker, Kotus and Lü2].

Set $r_n=M2^n$ for a large $M>0$ . For a given real number $\lambda>\tfrac 12$ , define $r_{n,k}=r_{n+1}-1+{k}/{[r_n^{\lambda }]},\ 1\leq k\leq [r_n^{\lambda }]$ for each n, where $[x]$ is the maximal integer not greater than x. We prove that the function

$$ \begin{align*}g(z)=\sum_{n=1}^{\infty}\frac{1}{[r_n^{\lambda}]}\sum_{k=1}^{[r_n^{\lambda}]}\frac{2z}{r_{n,k}^2-z^2} =\sum^{\infty}_{n=1}\frac{1}{[r_n^{\lambda}]}\sum_{k=1}^{[r_n^{\lambda}]}\bigg(\frac{1}{r_{n,k}-z}-\frac{1}{r_{n,k}+z}\bigg)\end{align*} $$

satisfies the requirement of Theorem 1.8.

Given an arbitrarily large real number r, we have $r_n\leq r<r_{n+1}$ for some n. Then,

$$ \begin{align*} n(r,f)&=\sum_{k=1}^{n-1}[r_k^{\lambda}]\leq r_n^{\lambda}\sum_{k=1}^{n-1}\bigg(\frac{r_k}{r_n}\bigg)^{\lambda}\\ &=r^{\lambda}\sum_{k=1}^{n-1}\bigg(\frac{1}{2^{n-k}}\bigg)^{\lambda}\\ &\leq \frac{r^{\lambda}}{2^{\lambda}-1} \end{align*} $$

and

$$ \begin{align*} n(r,f)&\geq \sum_{k=1}^{n-1}(r_k^{\lambda}-1)=\frac{r_n^{\lambda}}{2^{\lambda}-1}-n\\ &= \frac{r_{n+1}^{\lambda}}{2^{\lambda}(2^{\lambda}-1)}-\frac{\log(r_n/M)}{\log 2}\\ &> \frac{r^{\lambda}}{2^{\lambda}(2^{\lambda}-1)}-\frac{\log(r/M)}{\log 2}. \end{align*} $$

For $z\in A(\tfrac 43r_n,\tfrac 53r_n)$ ,

$$ \begin{align*} |g(z)|\leq\sum_{m=1}^{n-1}\frac{1}{[r_m^{\lambda}]} \sum_{k=1}^{[r_m^{\lambda}]}\frac{2|z|}{|z|^2-r_{m,k}^2}+ \sum_{m=n}^{\infty}\frac{1}{[r_m^{\lambda}]}\sum_{k=1}^{[r_m^{\lambda}]}\frac{2|z|}{r_{m,k}^2-|z|^2} =I_1+I_2(\text{say}). \end{align*} $$

We estimate

$$ \begin{align*} I_1&\leq \sum_{m=1}^{n-1}\frac{1}{[r_m^{\lambda}]}\sum_{k=1}^{[r_m^{\lambda}]} \frac{30r_n}{16r_n^2-9r_{m,k}^2}\\ &\leq\frac{30nr_n}{16r_n^2-9r_n^2}=\frac{30n}{7r_n}. \end{align*} $$

Since $r_{m,k}\geq r_{m+1}-1=2r_m-1$ , for $m\geq n$ ,

$$ \begin{align*} 9r^2_{m,k}-25r_n^2\geq 9(2r_m-1)^2-25r_n^2>10r_m^2 \end{align*} $$

so that

$$ \begin{align*} I_2&\leq \sum_{m=n}^{\infty}\frac{1}{[r_m^{\lambda}]}\sum_{k=1}^{[r_m^{\lambda}]} \frac{30r_n}{9r_{m,k}^2-25r_n^2}\\ &< \sum_{m=n}^{\infty}\frac{30r_n}{10r_{m}^2}\leq 3r_n\sum_{m=n}^{\infty}r_m^{-2}=\frac{3}{r_n}\sum_{m=n}^{\infty}\bigg(\frac{r_n}{r_m}\bigg)^{2}<\frac{3}{r_n}. \end{align*} $$

Therefore,

$$ \begin{align*} T(3r_n/2,g)=N(3r_n/2,g)+m(3r_n/2,g)<n(3r_n/2,g)\log(3r_n/2)+2 \end{align*} $$

and

$$ \begin{align*} T(3r_n/2,g)=N(3r_n/2,g)+m(3r_n/2,g)>n(r_n,g)\log(3/2)-2, \end{align*} $$

and so

$$ \begin{align*} \lim_{n\to\infty}\frac{\log T(3r_n/2,g)}{\log (3r_n/2)}=\lambda. \end{align*} $$

This easily implies that g has order and lower order equal to $\lambda $ .

Obviously, $0$ is an attracting fixed point of g and we can choose a large M such that $B(0,M)$ is in its attracting basin. For all sufficiently large n, the annulus $A(4r_n/3,5r_n/3)$ is mapped into $B(0,M)$ . Therefore, the Fatou set $F(g)$ contains a sequence of annuli $A(4r_n/3,5r_n/3)$ with $r_n\to \infty\ (n\to \infty )$ . Since $J(g)$ lies in the real axis and $r_1,r_2\in J(g)$ , g cannot take on the values $r_1$ and $r_2$ on the upper half plane and lower half plane.

We omit the proof of the case when $\lambda =\infty $ .